Java Solution in O(n)

// where n is the number of nodes in the binary tree
public static void topView(Node root) {
    int MAX = 500;    // Otherwise, MAX=getNodeCount(root);
    int[] top = new int[MAX*2];
    Queue<Object[]> queue = new ArrayDeque<>();
    queue.add(new Object[]{root, MAX});
    while(!queue.isEmpty()) {
        Object[] array = queue.remove();
        Node node = (Node)array[0];
        Integer order = (Integer)array[1];
        if(node == null) continue;

        if(top[order] == 0) top[order] = node.data;
        queue.add(new Object[]{node.left, order-1});
        queue.add(new Object[]{node.right, order+1});
    }

    for(int data: top) if(data != 0) System.out.print(data + " ");
}